When charcoal reacts in the presence of oxygen, CO and CO_2 are produced according to the following chemical reactions.
$$C(s) + \frac{1}{2}
O_{2}(g)
\rightarrow CO(g)$$
$$C(s) + O_{2}
(g) \rightarrow CO_{2}(g)$$

What would be the total mass of gas produced when $400$ g of charcoal is reacted, assuming equal amounts are consumed in each reaction.

Question

When charcoal reacts in the presence of oxygen, CO and CO_2 are produced according to the following chemical reactions.
$$C(s) + \frac{1}{2}
 O_{2}(g)
 \rightarrow CO(g)$$
$$C(s) + O_{2}
 (g) \rightarrow CO_{2}(g)$$

What would be the total mass of gas produced when $400$ g of charcoal is reacted, assuming equal amounts are consumed in each reaction.

anonymous · Answer

How am i able to solve this? Anyone able to give me a hint? What formula do i use?

australopithecus · Answer

well find the moles of carbon first

anonymous · Answer

mole of carbon is 12.01

australopithecus · Answer

now what is the limiting reagent in the initial reaction?

anonymous · Answer

what is the limiting reagent? Sorry my chemistry is really rusty..

australopithecus · Answer

In a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

anonymous · Answer

how am i able to calculate the limiting reagent then?

australopithecus · Answer

https://en.wikipedia.org/wiki/Limiting_reagent

australopithecus · Answer

it has been awhile since I have had to determine the limiting reagent let me refresh my memory but otherwise this problem is easy for me to solve :)

australopithecus · Answer

nevermind the limiting reagent is the charcoal

australopithecus · Answer

as it is what is going to be used up first

australopithecus · Answer

so once you have the moles you need to determine the amount of product since it is 1 to 1 you will have the same amount of moles of charcoal as you will for CO

australopithecus · Answer

the amount of moles of CO will equal the amount of moles of CO2 as it is still a 1 to 1 reaction

thus all you have to do is take the moles and use the formula

moles = grams/molecular mass

find the molecular mass of CO2 using the periodic table and solve for grams

anonymous · Answer

$$n=\frac{400}{44.01}$$

australopithecus · Answer

wait you are incorrect

australopithecus · Answer

let me work through this you gave me the molecular weight not the moles

anonymous · Answer

i know..since i got it wrong :/

anonymous · Answer

okay thanks

australopithecus · Answer

Write down the molecular mass

molecular mass of carbon is 12.01g/mol
molecular mass of oxygen is 15.9994g/mol
Thus the molcular mass of O2 is 15.9994*2 = 31.9988g/mol
Molecular mass of CO = 28.094
Molecular mass of CO2 = 12.01 + 15.9994*2 = 44.0088g/mol

Now we use the formula MEMORIZE THIS! 
moles = grams/molecular mass

so we need to find the moles of carbon we know that the charcoal weighs 400g

thus we have

400g/12.01g/mol = 33.3056mol of carbon

all the reactions are 1 to 1 thus we end up with (for every 1 mole of carbon you have 1 mol of product) so

33.3056mol = x/44.0088g/mol
=
1,465.7g

This is assuming that the amount of oxygen present is unlimited.

If the amount of oxygen was limited to 400g than you would need to find the limiting reagent

anonymous · Answer

the answer is $1.2$ kg..

australopithecus · Answer

Ok then I'm going to assume there are 400 grams of oxygen sorry the question isnt very clear

australopithecus · Answer

No I'm sorry but my answer has to be correct

400g/12.01g/mol = 33.30mol

33.30mol* 44.0088g/mol = 1466g of CO2

I will get someone to check it for me give me a second

anonymous · Answer

well the answer i have is $1.2$kg and its a multiple choice and the choices are 0.93, 1.5, 2.5 which does not much what you have..

australopithecus · Answer

wel 1.5 matches

australopithecus · Answer

if you round up

anonymous · Answer

but that is apparently not the right answer..

australopithecus · Answer

oh crud

australopithecus · Answer

oh I miss read the question equal amounts used in both reactions :L

australopithecus · Answer

200g in one reaction and 200g in the other

australopithecus · Answer

so 200g/12.01g/mol = 16.65mol

16.65mol*28.094 = 467.8g

then we have

200g/12.01g/mol = 16.65mol

16.65mol*44g/mol =732.87g

thus

467.8g + 732.87g = 1.2kg

australopithecus · Answer

there is your answer sorry for not understanding the question I have really bad insomnia lol

anonymous · Answer

yes im just looking over what you all solved, in essence when you calculate the moles of charcoal you halve that value and then recalculate to products because from stechiometry equal amount of charcoal burns to form CO and CO2...

anonymous · Answer

thank you!

australopithecus · Answer

man I feel like an idiot not comprehending the question and wasting my time ugh lol